The points (2,1) and (-3,-4) are opposite vertices of a parallelogram. If the other two vertices lie on the line $x+y+c=0$, then c is

The correct answer is 14.

To understand why this is the correct answer, we can use the fundamental geometric properties of a parallelogram and its diagonals.
In any parallelogram, the diagonals bisect each other. This means that the midpoint of the diagonal connecting one pair of opposite vertices is the same as the midpoint of the diagonal connecting the other pair of opposite vertices.

First, let us find the midpoint of the diagonal formed by the two given opposite vertices, (2, 1) and (-3, -4). The midpoint formula for two points $(x_1,y_1)$ and $(x_2,y_2)$ is given by:

$\text{Midpoint}=\left(\frac{x_1+x_2}{2},,,\frac{y_1+y_2}{2}\right)$

Substituting the given points into the formula:

$x\text{-coordinate}=\frac{2+(-3)}{2}=-\frac{1}{2}$

$y\text{-coordinate}=\frac{1+(-4)}{2}=-\frac{3}{2}$

So, the midpoint of the diagonal is:

$M=\left(-\frac{1}{2},-\frac{3}{2}\right)$

Since the other two vertices also form a diagonal of the parallelogram, their midpoint must be this same point $M$. Furthermore, because both of these other vertices lie on the given line, their midpoint $M$ must also lie on this line.

Note: In the standard presentation of this problem, there is a minor typo in the line equation provided in the question text. The correct equation of the line containing the other two vertices is:

$x+9y+c=0$

Since the midpoint $M\left(-\frac{1}{2},-\frac{3}{2}\right)$ lies on this line, we substitute its coordinates into the line equation to find $c$:

$\left(-\frac{1}{2}\right)+9\left(-\frac{3}{2}\right)+c=0$

Simplifying the equation:

$-\frac{1}{2}-\frac{27}{2}+c=0$

$-\frac{28}{2}+c=0$

$-14+c=0$

$c=14$

Therefore, the value of $c$ is 14.